-1
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$\lim_{x \to 2} \frac{3x^2-13}{x-2}$ How can this be done?Putting $x=2$ also doesn't give us $\frac{0}{0}$ form for which we could apply L'hopitals rule.Any ideas?

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  • $\begingroup$ isn't this just infinity? Maybe you wanted to write -12 then it becomes $3(x^2-4)$ which can be written as $3(x-2)(x+2)$ $\endgroup$ Jul 15 at 19:29
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$\lim_{x\rightarrow2^+}\frac{3x^2-13}{x-2}=\frac{-1}{0^+}=-\infty$ and $\lim_{x\rightarrow2^-}\frac{3x^2-13}{x-2}=\frac{-1}{0^-}=\infty$. So this limit doesn't exist.

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  • $\begingroup$ I downvoted this answer because $\dfrac{-1}{0}$ is not defined. If $f(x)\to-1$ and $g(x)\to0$ as $x \to a^+$, and there is a $\delta>0$ such that if $0<x-a<\delta$ then $g(x)$ is positive, then $\dfrac{f(x)}{g(x)}\to\infty$ as $x\to a^+$; simply saying that $\dfrac{-1}{0}=-\infty$ is too sloppy in my opinion. $\endgroup$
    – Joe
    Jul 15 at 21:42
  • $\begingroup$ @Joe Yes. You are right. Rigorously, we should use $\epsilon$-$\delta$ language. However, this is an intuitive result. When f(a) is not zero and g(a) is zero, you can always use this technic to see the limit quickly. $\endgroup$
    – Zhiyan
    Jul 17 at 1:29
  • $\begingroup$ I edited your post a little to make it clearer that we are not dividing by zero, and I have removed my downvote. $\endgroup$
    – Joe
    Jul 17 at 20:47

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